Balancing Chemical Equation

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Balancing chemical equation:<br/>In this you will be learning how to balance a chemical equation, step by step instructions are given , Three examples are worked out ,and shown how it can be balanced, students will learn simple steps to balance the chemical equation

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Balancing the chemical equations : Balancing the chemical equations

Steps to Balance chemical equations : Steps to Balance chemical equations 1. Write down the unbalanced chemical equation. 2.Next note down the number of elements present on each side of the equation 3.Count the number of molecules present on each side of the equation. 4.Check weather there are same number of molecules of all elements are present on the reactants side as well as the products side 5. start with the element that has the largest number of atoms in a single compound 6. Add coefficients to the element or ion that needs to be increased . 7. Start by balancing one element at a time. Finally check if all the elements are balanced.

Examples : Examples 1) Al + O2 Al2O3 Count atoms on both side of equation Left hand side Right hand side Al = 1 Al =2 O = 2 O =3 Lets start with Aluminum (Al) ,since there is 1 Al on left side and 2 Al on right side multiply Al by 2 on left side. 2Al + O2 Al2O3 Now lets look at the Oxygen (O), we have 2 Oxygen on left side and 3 Oxygen on right side. Multiply Oxygen by 1.5 on left side to make it 3 2Al +1.5 O2 Al2O3 We cannot have 0.5 of an atom or 1.5 , so we multiply the whole equation by 2, the final balanced equation we get is: 4Al + 3 O2 2Al2O3

Slide 4 : C2H6 + O2 CO2 + H2O Left hand side Right hand side C=2 C=1 H=6 H=2 O=2 O=3 First lets start with Carbon ( C) , we have 2 Carbon on left hand side and 1 Carbon on right hand side, so we multiply C by 2 on right hand side of the equation. C2H6 + O2 2CO2 + H2O Next coming on to Hydrogen (H), We have 6 hydrogen on left side and 2 hydrogen on the right side, we multiply H by 3 on right side to make it 6 C2H6 + O2 2CO2 + 3H2O Now lets look at Oxygen (O) , we have 2 Oxygen on left side, 7 Oxygen on right side. So multiply Oxygen by 3.5 left side to make it 7 C2H6 + 3.5O2 2CO2 + 3H2O Since we cant have fractions, multiply the whole equation by 2 , we get the balanced equation 2C2H6 + 7O2 4CO2 + 6H2O

Slide 5 : 3) Fe2O3 +H2SO4 Fe2(SO4)3+ H2O In the above equation the sulphate (SO4) stays together , so we can replace SO4 = X, we can rewrite the above equation as: Fe2O3 +H2X Fe2X3+ H2O Left hand side Right hand side Fe =2 Fe =2 O =3 O =1 H=2 H=2 X=1 X=3 Let us look at Fe , it is 2 on both sides so its balanced Next let us look at O , its 3 on left side and 1 on right side , so we multiply O By 3 right side Fe2O3 +H2X Fe2X3+ 3H2O Next lets look at H , we have 2 on left side and 6 on right side , so we multiply H by 3 on left side Fe2O3 +3H2X Fe2X3+ 3H2O Now lets look at X , it 3 on both the sides so X is balanced, finally we can replace X =SO4 we write the balanced equation as: Fe2O3 +3H2SO4 Fe2(SO4)3+ 3H2O

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Slide 7 : Thank You

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